a) Ta có: \(a^3y^3+125\)

\(=\left(ay+5\right)\left(a^2y^2-5ay+25\right)\)

b) Ta có: \(8x^3-y^3-6xy\cdot\left(2x-y\right)\)

\(=\left(2x-y\right)\left(4x^2+2xy+y^2\right)-6xy\left(2x-y\right)\)

\(=\left(2x-y\right)\left(4x^2+2xy-6xy+y^2\right)\)

\(=\left(2x-y\right)^3\)

\(\left(x-y\right)^3+\left(x+y\right)^3\\ =\left(x-y+x+y\right)\left(\left(x-y\right)^2-\left(x-y\right)\left(x+y\right)+\left(x+y\right)^2\right)\\ =2x\left(x^2-2xy+y^2-\left(x^2-y^2\right)+x^2+2xy+y^2\right)\\ =2x\left(x^2-2xy+y^2-x^2+y^2+x^2+2xy+y^2\right)\\ =2x\left(x^2+3y^2\right)\)

\(\left(x-y\right)^3+\left(x+y\right)^3\)

\(=\left(x-y+x+y\right)\left[\left(x-y\right)^2-\left(x-y\right)\left(x+y\right)+\left(x+y\right)^2\right]\)

\(=2x\left(x^2-2xy+y^2-x^2+y^2+x^2+2xy+y^2\right)\)

\(=2x\left(x^2+3y^2\right)\)

\(\left(x-y\right)^3-\left(x+y\right)^3\\ =\left(x-y-x-y\right)\left(\left(x-y\right)^2+\left(x-y\right)\left(x+y\right)+\left(x+y\right)^2\right)\\ =-2y\left(x^2-2xy+y^2+x^2-y^2+x^2+2xy+y^2\right)\\ =-2y\left(3x^2+y^2\right)\)

Các chổ này chị dùng ngoặc vuông nha 

 =(y+x-6)(y+x-2)

Đặt \(x+y=u\)

Biểu thức trở thành \(u^2-8u+12\)

\(=u^2-2u-6u+12\)

\(=u\left(u-2\right)-6\left(u-2\right)\)

\(=\left(u-6\right)\left(u-2\right)\)

Thay ngược trở lại, ta được:

\(\left(x+y\right)^2-8\left(x+y\right)+12=\left(x+y-6\right)\left(x+y-2\right)\)

Trả lời:

1, 15x + 15y = 15 ( x + y )

2, 8x - 12y = 4 ( 2x - 3y )

3, xy - x = x ( y - 1 )

4, x² + x = x ( x + 1 )

5, 3x²y - 8xy² = xy ( 3x - 8y )

6, 6x - 12xy - 18x² = 6x ( 1 - 2y - 3x )

1) 15x + 15y = 15(x + y)

2) 8x - 12y = 4(2x - 3y)

3) xy - x = x(y - 1)

4) x² + x = x(x + 1) 

5) 3x²y - 8xy² = xy(3x - 8y)

6) 6x - 12xy - 18x² = 6x(1 - 2y - 3x) 

\(x^2+2xy+7x+7y+y^{2+10}\)

\(\text{phân tích đa thức thành nhân tử}\)

\(y^{12}+2xy+7y+x^2+7x\)

tách \(^{x^2}\)ra rồi làm thừa số chung, toán SGK đem ra hỏi làm j

\(x^2-y^2+8x+6y+7\)

\(=\left(x-y\right)\left(x+y\right)+7\left(x+y\right)+x-y+7\)

\(=\left(x+y\right)\left(x-y+7\right)+\left(x-y+7\right)\)

\(=\left(x+y+1\right)\left(x-y+7\right)\)

Ta có x² - y² + 8x + 6y + 7

= x² + 8x + 16 - y² + 6y - 9

= \(x^2+4x+4x+16-y^2+3y+3y-9\)

= x(x + 4) + 4(x + 4) - y(y - 3) + 3(y - 3)

= (x + 4)² - (y - 3)²

= (x + 4 + y - 3)(x + 4 - y + 3) 

= (x + y + 1)(x - y + 7)

=(x-y-2y)[(x-y)^2+2y(x-y)+4y^2]

=(x-3y)(x^2-2xy+y^2+2xy-2y^2+4y^2)

=(x-3y)(x^2+3y^2)

\(\left(x-y\right)^3-8y^3\)

\(=\left(x-y\right)^3-\left(2y\right)^3\)

\(=\left[\left(x-y\right)-2y\right]\left[\left(x-y\right)^2+2y\left(x-y\right)+\left(2y\right)^2\right]\)

\(=\left(x-y-2y\right)\left(x^2-2xy+y^2+2xy-2y^2+4y^2\right)\)

\(=\left(x-3y\right)\left(x^2+3y^2\right)\)

\(x^2-y^2\)

\(=x^2+xy-xy-y^2\)

\(=x\left(x+y\right)-y\left(x+y\right)\)

\(=\left(x+y\right)\left(x-y\right)\)

x²-y²=(x+y)(x-y)